Subgroup generated by $1 - \sqrt{2}$, $2 - \sqrt{3}$, $\sqrt{3} - \sqrt{2}$ For a number field $K$, Dirichlet's unit theorem says that$$(O_K)^\times = \mathbb{Z}^{r - 1} \oplus (\text{a finite cyclic group}),$$where $r$ is the number of all infinite places of $K$. An infinite place of $K$ is either an embedding of $K$ (as a field) into $\mathbb{R}$ or the complex conjugacy class of an embedding of $K$ (as a field) into $\mathbb{C}$ with dense image.
When $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, how do I see that $1 - \sqrt{2}$, $2 - \sqrt{3}$, and $\sqrt{3} - \sqrt{2}$ generate a subgroup $\Gamma$ of $(O_K)^\times$ such that$$\Gamma \cong \mathbb{Z}^3 = \mathbb{Z}^{r - 1}?$$
 A: Recall the proof of Dirichlet's unit theorem uses the logarithmic embedding

$$l:\begin{cases}\mathcal{O}_K^\times\to \Bbb R^{r+s} \\
\epsilon\mapsto (\log|\sigma_1(\epsilon)|,\ldots , \log|\sigma_r(\epsilon)|,\log|\sigma_{r+1}(\epsilon)|,\ldots \log|\sigma_{r+s}(\epsilon)|)\end{cases}$$

where the first $r$ are the real embeddings and the second $s$ are representatives of complex embeddings. In this case $r=4$ and $s=0$. Then Dirichlet's unit theorem asserts that the only linear forms which annihilate all such log vectors on units of $\mathcal{O}_K$ are multiples of the form represented by $(1,1,1,1)$. Our strategy will to look at other forms that annihilate individual log vectors and to show they generate different dual spaces, which is exactly equivalent to showing linear independence.
Now consider that each of your three units--which I will name $\epsilon_1, \epsilon_2,\epsilon_3$--generate different fields, the first two generate two of the quadratic subfields, and the third generates the entire field, $K$. Let $G_K$ be the Galois group of $K$ generated by $\sigma,\tau$ where $\sigma(\sqrt{2})=-\sqrt{2}$ and fixes $\sqrt 3$ and $\tau(\sqrt{3})=-\sqrt 3$ and fixes $\sqrt 2$.
Then we have that

$$\begin{cases} l(\epsilon_1) = (\log|\epsilon_1|, \log|\sigma(\epsilon_1)|, \log|\epsilon_1|, \log|\sigma(\epsilon_1)|)\\
l(\epsilon_2) = (\log|\epsilon_2|, \log|\epsilon_2|, \log|\tau(\epsilon_2)|, \log|\tau(\epsilon_2)|) \\
l(\epsilon_3) = (\log|\epsilon_3|, \log|\sigma(\epsilon_3)|, \log|\tau(\epsilon_3)|, \log|\sigma\tau(\epsilon_3)|)
\end{cases}.$$

The first vector is annihilated by the forms represented by $(1,1,0,0)$ and $(1,0,0,1)$, together with $(1,1,1,1)$ this uniquely determines the linear subspace in which it is contained. Similarly we see the second log vector is annhilated by $(1,0,1,0)$ and $(0,1,1,0)$ in addition to the one they all share. But then since these have different dual spaces, they are not linearly dependent, hence they generate a $2$-dimensional subspace of $\Bbb R^4$. We must now show that $l(\epsilon_3)$ is not annihilated by the intersection of the two dual spaces. We compute (using column vectors for typesetting ease, but all are secretly their transposes as we will use in the sequel) to see that
$$l(\epsilon_1)^*\cap l(\epsilon_2)^* = \operatorname{span}\left\{\begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}1 \\ 0 \\ 0 \\ 1\end{pmatrix}, \begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}\right\}\bigcap\operatorname{span}\left\{\begin{pmatrix}1 \\ 0 \\ 1 \\ 0\end{pmatrix},\begin{pmatrix}0 \\ 1 \\ 1 \\ 0\end{pmatrix},\begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}\right\}$$
It is enough to find a vector not a multiple of $\begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}$ since clearly that is in the intersection. We setup the system of equations:

$$a\begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}+b\begin{pmatrix}1 \\ 0 \\ 0 \\ 1\end{pmatrix}+c\begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix} = \alpha \begin{pmatrix}1 \\ 0 \\ 1 \\ 0\end{pmatrix}+\beta \begin{pmatrix}0 \\ 1 \\ 1 \\ 0\end{pmatrix}$$

and by inspection $\beta = c=-b$ gives a solution, so the intersection is
$$\operatorname{span}\left\{\begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}, \begin{pmatrix}1 \\ 0 \\ 0 \\ 1\end{pmatrix}\right\}$$
Only the first form annihilates $l(\epsilon_3)$, hence $l(\epsilon_3)$ is not in $\operatorname{span}\{l(\epsilon_1),l(\epsilon_2)\}$, proving the assertion.
A: Heuristically, each of the units comes from a different quadratic subextension of $K$ - e.g. $\sqrt2-1$ is a fundamental unit in $\mathbb Q(\sqrt 2)$, and $5-2\sqrt6 = (\sqrt2-\sqrt3)^2$ is a fundamental unit in $\mathbb Q(\sqrt 6)$. We can exploit this as follows:
Following @GregMartin's comment, assume that $(\sqrt2-1)^a(2-\sqrt3)^b(\sqrt3-\sqrt2)^c=1$; we need to show that $a=b=c=0$.
Suppose, for example, that $a\ne 0$. Since $1-\sqrt2$ is a fundamental unit of $\mathbb Q(\sqrt2)$, it follows that $b$ and $c$ cannot both be $0$. So
$$(\sqrt 2-1)^{-a} = (1+\sqrt2)^a=(2-\sqrt3)^b(\sqrt3-\sqrt2)^c.$$
Since $2-\sqrt3$ and $(\sqrt3-\sqrt2)^2$ are fundamental units of $\mathbb Q(\sqrt3)$ and $\mathbb Q(\sqrt6)$, one can now check that regardless of which of $b,c$ is non-zero, the right hand side will have non-trivial terms in $\sqrt3$ or $\sqrt6$, whereas the left hand side cannot.
A similar argument rules out the cases where $b,c\ne 0$.
A: This amounts to showing that $1-\sqrt2,2-\sqrt3,\sqrt3-\sqrt2$ are multiplicatively independent. Suppose for some non-zero integers $\alpha,\beta,\gamma$ we have that $(1-\sqrt2)^\alpha(2-\sqrt3)^\beta=(\sqrt3-\sqrt2)^\gamma$. Squaring this relation we get 
$$(1-\sqrt2)^{2\alpha}(2-\sqrt3)^{2\beta}=(5-2\sqrt 6)^\gamma. $$
Let $(1-\sqrt2)^{2\alpha}=a+b\sqrt 2$ and $(2-\sqrt3)^{2\beta}=c+d\sqrt 3$ and $(5-2\sqrt 6)^\gamma=e+f\sqrt 6$. Now $e$ cannot be zero for else $f\sqrt 6$ will be a unit, an impossibility. Now $f$ cannot be zero either because then $e$ would be a unit, i.e. $\pm 1$, which is absurd since every power of $|5-2\sqrt 6|<1$ and hence no non-zero power of this can be $1$.
If $$(a+b\sqrt2)(c+d\sqrt3)=ac+ad\sqrt3+bc\sqrt2+bd\sqrt6=(e+f\sqrt6),$$
then neither $b$ nor $d$ can be zero so it must be the case that both $a=c=0$. But in this case $ac=e=0$ but $e$ cannot be zero. So there do not exist such non-zero integers $\alpha, \beta,\gamma$.
A: Expanding what I believe Franz Lemmermeyer meant with his hint. This is closely built into all the other answers as well (+1 to y'all), but I learned a trick from reading all this, so I can't keep the lid on...
Assume contrariwise that there exist a non-trivial triple $(a,b,c)$ of integers such that
$$
|(1-\sqrt2)^a(2-\sqrt3)^b(\sqrt3-\sqrt2)^c|=1.\qquad(i)
$$
I use the absolute value signs so that I don't need to worry about the torsion unit. Let's apply the automorphism $\tau:\sqrt2\mapsto\sqrt2,\sqrt3\mapsto-\sqrt3$ to equation $(i)$. As $\tau(1-\sqrt2)=(1-\sqrt2)$, $\tau(2-\sqrt3)=(2-\sqrt3)^{-1}$ and $\tau(\sqrt3-\sqrt2)=-(\sqrt3-\sqrt2)^{-1}$, we get another equation
$$
|(1-\sqrt2)^a(2-\sqrt3)^{-b}(\sqrt3-\sqrt2)^{-c}|=1.\qquad(ii)
$$
Multiplying $(i)$ and $(ii)$ together gives
$$
|(1-\sqrt2)^{2a}|=1\qquad(iii)
$$
from which we can infer that $a=0$.
There are several ways to continue. By using another automorphism $\sigma:\sqrt2\mapsto-\sqrt2, \sqrt3\mapsto-\sqrt3$ we similarly reach the conclusion $c=0$. Continuing in the same vein, using the automorphism $\sigma\tau$ gives us $b=0$.
Essentially we are taking the relative norms of these units to the three quadratic subfields and checking what that gives us.
Alternatively, after having derived $(iii)$ we can use Mathmo123's idea. The equations $(i)$ and $(iii)$ imply that $(2-\sqrt3)^{2b}$ and $(\sqrt3-\sqrt2)^{2c}$ are inverses of each other. But the former is an element of $\Bbb{Q}(\sqrt3)$ and the latter of $\Bbb{Q}(\sqrt6)$. The intersection of the two fields is $\Bbb{Q}$, so both must be rational. But the only rational units are $\pm1$, so this is absurd.
